= Solution
Away from a bifurcation set, the normal to a <Killing horizon> is its nonzero <Killing vector field> $K$. Any affine tangent to the same generators has the form $k=hK$. In the <optical tensor>, derivatives of $h$ contribute a factor of $K_a$ and disappear under the <screen-space projector>. Consequently
$$
B_{(ab)}=h\,q_a{}^cq_b{}^d\nabla_{(d}K_{c)}=0
$$
by the <Killing equation>. The trace and trace-free symmetric part are therefore zero: $\theta=0$ and $\sigma_{ab}=0$.
For the antisymmetric part, locally the horizon normal is proportional to a gradient. The restriction of its dual one-form to every tangent direction on the horizon is zero. Taking its <exterior derivative> and evaluating on two screen tangents therefore gives zero as well; equivalently this is the hypersurface-orthogonality consequence of the <Frobenius theorem>. Since both screen indices are tangential, $B_{[ab]}=0$. Thus the <optical scalars on a Killing horizon> satisfy
$$
\boxed{\theta=0,\qquad\sigma_{ab}=0,\qquad\omega_{ab}=0.}
$$
No affine normalization of the <Killing vector field> itself was assumed. At a regular <bifurcation surface>, $K$ vanishes but the smooth null generators do not; the conclusions for their optical quantities follow by continuity along each horizon branch.
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